How do you solve $sec^2x-2=0$ ?

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Answer 1 · 2016-09-16T14:57:28

The general solution is $npi+-pi/3$

As $sec^2x-2=0$ , we have

$sec^2x-(sqrt2)^2=0$ or

$(secx+sqrt2)(secx-sqrt2)=0$ and

$secx=+-sqrt2=+-sec(pi/3)$

Hence in the range $(-pi,+pi)$ the values are ${+-pi/3,+-(2pi)/3}$ , which can also be written as ${+-pi/3,+-(pi-pi/3)}$

Hence, the general solution is $npi+-pi/3$

How do you solve sec^2x-2=0sec^2x-2=0?